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Math Help - calculus - line integral and work done

  1. #1
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    Question calculus - line integral and work done

    question 1 -line integral

    Evaluate the integral of C xdy-ydx along the curve C given by the equation y=x^3 from (0,0) to (2,8).


    question 2 - work

    Find the work done in moving an object through the vector field
    F<x,y> = <-y,x> along the upper semicircle x^2 + y^2 =1 from A(1,0) to B(-1,0).

    Please HELP ! Thank you very much.
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by kittycat View Post
    question 1 -line integral

    Evaluate the integral of C xdy-ydx along the curve C given by the equation y=x^3 from (0,0) to (2,8).
    you can parametrize.. let x=t so y=t^3, then the limits of t will be from 0 to 2.. so, dx=dt and dy=3t^2 dt. therefore
     xdy-ydx = 3t^3 dt - t^3 dt = 2t^3 dt \, so,
    \int_0^2 {2t^3 \, dt} = \frac{1}{2}t^4 evaluated from 0 to 2 equals 8..

    Quote Originally Posted by kittycat View Post
    question 2 - work

    Find the work done in moving an object through the vector field
    F<x,y> = <-y,x> along the upper semicircle x^2 + y^2 =1 from A(1,0) to B(-1,0).
    so you have F(x,y) = -yi + xj

    let x=t,  y= \sqrt{1-t^2}, \, -1 \leq t \leq 1

    so,

    Work = \int_C Fdr = \int_C {-ydx + xdy} = \int_{-1}^1 {-\sqrt{1-t^2} dt + t \frac{-t}{\sqrt{1-t^2}}dt}

     = \int_{-1}^1 \frac {t^2-1 -t^2}{\sqrt{1-t^2}}dt = \int_{-1}^1 \frac {-dt}{\sqrt{1-t^2}}

    i think, you can do the rest..
    Last edited by kalagota; October 30th 2007 at 11:01 PM.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by kalagota View Post
    you can parametrize.. let x=t so y=t^3, then the limits of t will be from 0 to 2.. so, dx=dt and dy=3t^2 dt. therefore
     xdy-ydx = 3t^3 dt - t^3 dt = 2t^3 dt \, so,
    \int_0^2 {2t^3 \, dt} = \frac{1}{2}t^4 evaluated from 0 to 2 equals 8..



    so you have F(x,y) = -yi + xj

    let x=t,  y= \sqrt{1-t^2}, \, -1 \leq t \leq 1

    so,

    Work = \int_C Fdr = \int_C {-ydx + xdy} = \int_{-1}^1 {-\sqrt{1-t^2} dt + t \frac{-t}{\sqrt{1-t^2}}dt}

     = \int_{-1}^1 \frac {t^2-1 -t^2}{\sqrt{1-t^2}}dt = \int_{-1}^1 \frac {-dt}{\sqrt{1-t^2}}

    i think, you can do the rest..
    This is correct, but a detail has been omitted. The definition of work is
    W = \int_C F \cdot ds
    not
    Work = \int_C Fdr
    (The missing dot product is incorrect, and the dr implies a radial differential.)

    So
    W = \int_C F \cdot ds = \int_C (-y \hat{i} + x \hat{j}) \cdot (dx \hat{i} + dy \hat{j}) = \int_C -ydx + xdy
    as kalagota stated.

    -Dan
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  4. #4
    MHF Contributor kalagota's Avatar
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    well, i just omitted the dot.. anyways, according to my notes (a year ago ), the definition states

    W= \int_C F \cdot Tds

    where T=\frac {dr}{ds}\, \, the tangential vector, that is why i came with

    \int_C F \cdot dr..

    anyways, thanks!!
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by kalagota View Post
    well, i just omitted the dot.. anyways, according to my notes (a year ago ), the definition states

    W= \int_C F \cdot Tds

    where T=\frac {dr}{ds}\, \, the tangential vector, that is why i came with

    \int_C F \cdot dr..

    anyways, thanks!!
    I do it that way because I'm a Physicist, not a Mathematician.

    Either way, the point is you get to the correct answer.

    -Dan
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  6. #6
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by topsquark View Post
    I do it that way because I'm a Physicist, not a Mathematician.

    Either way, the point is you get to the correct answer.

    -Dan
    hehe.. thanks!!
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    well, i just omitted the dot.. anyways, according to my notes (a year ago ), the definition states



    where the tangential vector, that is why i came with

    ..

    =================================

    Hi Kalagota,

    My lecture note states that T is a unit tangent vector where T(t) = r'(t)/IIr'(t)II .

    Does this mean the same as what you mentioned --- the tangential vector?
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  8. #8
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by kittycat View Post
    well, i just omitted the dot.. anyways, according to my notes (a year ago ), the definition states



    where the tangential vector, that is why i came with

    ..

    =================================

    Hi Kalagota,

    My lecture note states that T is a unit tangent vector where T(t) = r'(t)/IIr'(t)II .

    Does this mean the same as what you mentioned --- the tangential vector?
    yes, they are just the same. the reason is that, you gave the function in terms of the parameter t; while mine was parameter s with respect to the arc length..
    actually,
    ||r'(t)|| = s'(t)
    then from here, you will that they are the same..
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  9. #9
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    Question

    Hi Kalagota,

    Could you please tell me how to solve the following question? Thank you very much.

    question (work )
    Evaluate the integral c of F dot dr where F(x,y,z) = 8x^2yz i+ 5zj-4xyk and c is the curve given by r(t)=ti + t^2j+t^3k where 0 < or = t< or = 1.
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  10. #10
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by kittycat View Post
    Hi Kalagota,

    Could you please tell me how to solve the following question? Thank you very much.

    question (work )
    Evaluate the integral c of F dot dr where F(x,y,z) = 8x^2yz i+ 5zj-4xyk and c is the curve given by r(t)=ti + t^2j+t^3k where 0 < or = t< or = 1.
    recall, if \bold{F} is a continuous vector field on a smooth curve C given by the vector function \bold{r} (t), a \le t \le b. Then the line integral of \bold{F} along C is given by:

    \int_C \bold{F} \cdot d\bold{r} = \int_a^b \bold{F}( \bold{r}(t)) \cdot \bold{r}'(t)~dt = \int_C \bold{F} \cdot \bold{T}~ds

    use the middle one
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